 Overview of Section 2.3 More on Constructions

This section looks at problems posed about constructions, most of them making use of the Inscribed Angle Theorem (Theorem 2.2), its corollary, and interpretations. The essential idea is that an angle subtending and arc or its corresponding chord will have measure equal to 1/2 of the corresponding central. The central angle is the same regardless where the vertex of the angle is located on the arc of the circle.

The demonstrations in this section, elsewhere in the text, and in class emphasize three part of the problem solving process for constructions:

1. Investigation or analysis. The usual approach is to imagine the problem is solved and search for relationships or properties that will allow us to accomplish the construction.

2. Construction. We propose the steps in the construction and perform the construction.

3. Proof: We justify that the constructions steps are valid and that the construction accomplishes what it was supposed to do.

Construction 2.1

Construct the locus of all points from which a given segment a is seen at a given angle A.

Comment: This problem is asking us to show a construction given certain input: a segment and an angle. When we image the problem as solved, the inscribed angle theorem tells us that the vertex of angle A is on the arc of a circle and the angle subtends the segment as a chord of a circle. On the left is the arc with on case of the an angle A subtending the given segment.

Open GSP file to see an animation as the vertex is moved along the locus.

Now let's look for properties and relationships that can let os reach this construction. From the figure, the problem is to locate the center of the circle. First that point will lie on the perpendicular bisector of given line segment. Since the central angle is twice the size of the angle at the arc, the perpendicular bisector would divide the central angle into two angles, each the size of our given angle. Now a parallel line to our perpendicular bisector and one end of the segment, that is, a line perpendicular to the segment, would have an angle of the same size determined by the segment from the central angle as a transversal of the two parallel line.

Thus we could have the following construction:

1. Construct given segment a and label its ends B and C.

2. Construct the perpendicular bisector of BC

3. Construct a perpendicular to BC and B

4. Copy the given angle by construction to locate the vertex at B and one side along the perpendicular bisector at A.

5. The other side of the copied angle with intersect the perpendicular bisector of BC at O, the center of the circle.

6. Construct the circle and not the arc on the same side of BC as A.

Proof: By construction, the given angle has been located at B with one side along the perpendicular at B. The perpendicular bisector is parallel and the side of the angle is a transversal. Therefore half of the central angle is congruent to the given angle and the central angle is twice the given angle. Therefore any angle with a vertex on the circle will subtend the given chord and have an angle equal to the given angle.

Now Solve This 2.9

This NST shows an alternative construction for the locus of points given a segment and a subtending angle. Open GSP file. Essentially, the construction uses the given angle to construct a right triangle by constructing a perpendicular to one of the sides of the given angle, marking off the length of the given segment on the perpendicular, and then constructing a perpendicular at the other end of the segment to intersect with the second side of the triangle. From there, a perpendicular back to the first side of the given angle will complete a right triangle with the given chord as one of the legs. The circumcircle of this triangle will then provide the locus. Another alternative would be to pick an appropriate point on one side of the given angle, then strike an arc of the length of the given segment to determine one or two triangles that could be used to construct a circumcircle that will provide the locus circle.

Note that in all of these constructions for Now Solve This 2.9, the locus shown here is only 1/2 of the locus. The other part would be a reflection in the segment.

The constructions here were for an acute angle. They will also work for a right angle. Check it out.

If the given angle is an obtuse angle, then locus is a minor arc and the constructions need to be modified to use the supplement of the given angle.

PLEASE NOTE THE 'Notations for Construction Problems" on page 107.

Because of the difficult of subscripts, Greek letters, and symbols on the web, some translation of this notation will be adopted:

Triangle Construction Problems

In these problems, information is given about a triangle to identify physical attributes. The problem is to describe ruler and compass constructions of the triangle from the given attributes. For consistent notation adapted to this environment I use the following: The angles at vertices A, B, C: A, B, C
The sides of the triangle opposite vertices A, B, C: a, b, c.
The medians: m(a), m(b), m(c)
The altitudes: h(a), h(b), h(c)
The segments along an angle bisector to the opposite side: d(a), d(b), d(c)
The radius of the circumcircle: R
The radius of the incircle: r
The radii of the excircles: r(a), r(b), r(c)

The notation m(a), etc., is usually written with a subscript and I will do so in the sketches. The notation: b+c will mean "given a line segment equal in length to the sum of the lengths of sides b and c.

Construction 2.2

Construct a triangle from a, <A, h(a). That is, given a segment of length a for one side of the triangle (side BC), The angle A opposite the side, and the altitude to the side, construct the triangle. These are the given data.

This is the triangle we wish to construct.

Investigation: Angle A subtends the segment BC of length a so we could find one locus where the vertex A of the triangle could be located with the construction we did in Construction 2.1; it would be on the major arc of a particular circle determined by the given angle and the given segment.

A second locus for vertex A would be determined by BC and the altitude. The vertex A would have to lie along a line parallel BC.

Construction: 1. Copy segment a and label the ends B and C.

2. Construct Parallel to BC a distance h(a) away from it.

3. Construct the locus that is an arc of circle determined by BC and angle A.

4. When the parallel line intersects the circle would be the possible location of the vertex A satisfying both conditions. The two triangles are congruent.

OPEN GSP FILE

Now Solve This 2.10.

Prove the two triangles obtained in your construction are congruent.

Construction 2.3 Construct a triangle from a, A, m(b)

That is, construct a triangle given one side, the angle opposite that side, and the median to a second side.

Investigation: Imagine the problem is solved. The resulting triangle should look like the one at the right.

This is a bit like Problem 1.4.11 where the construction was for

a, b, m(c). In that problem we needed to locate an intermediate construction to arrive at the desired result and we made use of the midsegment theorem the that the line segment between the midpoints of two sides is parallel to the third side and half its length. So, the fact that MN is parallel to AB means angle at N is the same as at the vertex A. So we could locate N as the intersection of two loci. The first is on a circle of radius m(b) centered at B. The second is on the arc of a circle determine by angle A and MC, one-half of BC.

Construction: 1. Construct BC congruent to the given segment of length a

2. Find M, the midpoint of BC. Locate segment MC.

3. Construct a circle with center at B and radius m(b)

4. Construct locus determine by angle A and MC

5. Find N, an intersection of the two loci.

6. Extend CN to twice its length to locate point A

7. Construct AB.

Open GSP file?

Proof:

BC is congruent to the given segment by construction. MC is half of BC. Angle MNC is the given angle and because MN is parallel to AB, the angle at A is the given angle. CN = AN by construction and so BN is the median, but by construction BN was of length m(b).

Now Solve This 2.11. Alternative construction for a, A, m(b). Consider the desired triangle again but extend the median BN to twice its length to a point D and construct AD and CD

Why is this a parallelogram? Since ABCD is a parallelogram

Construction 2.4.

From a point P outside a circle, construct tangents to the circle with center O. This is one of the more fundamental constructions, so its derivation, execution, and proof should be understood by anyone who learns Euclidean Geometry. Investigation:
If we take the construction as completed, the line OP would create two congruent right triangles with hypotenuse OP, one leg a radius of the circle, and one leg a tangent. But a right triangle would have a circumcircle with the hypotenuse along the diameter. From the given information, we can construct the circle with diameter OP and its intersections with the given circle will form right triangles and hence locate the points of tangency. Construction:
1. Construct OP
2. Locate midpoint of OP
3. Construct circle with diameter OP
4. Construct intersection points K , L
of the circumcircle with the given
circle.
5. PK and PL are the tangents.

Proof:
By construction, Triangle OKP and Triangle OLP are right triangles with right angles at K and L.
OK and OL are radii of the circle.
Therefore, K and L are points of tangency.

EXTRA

Given a circle O and a point P outside the circle. Construct a circle tangent to the given circle and passing through P. That is, pick a point Q on the circle and construct a circle tangent to the given circle at Q and passing through P.

Make a GSP construction for this and when you are finished, move (animate) point Q around the circle. Trace the locus of the center of your tangent circle.

Discuss how the tangent lines in Construction 2.4 are limit cases when Q is at K or H. Describe.

Send me an e-mail with your result.

Victor

Jackie

Nicholas

Mine

Construction 2.5. Common Tangents (tangent lines) to two circles.

Given two circles O1 and O2, construct the common tangent lines to the two circles.

Note: There are four of them if the circles are disjoint: There are three of them if the two circles are tangent:

We know from Problem 2.2.14 that the internal tangent common to both circles is the perpendicular to the line of centers. There are two of them if the two circles intersect.

There are no common tangent lines if one circle is containing inside the other.

Investigation for Construction 2.5. All of these case can have a common strategy. We want to incorporate Construction 2.4 into our construction strategy here. For the external tangents in Case 1, 2 and 3, some auxiliary constructions will help.

We can construct a rectangle on the parallel radii for the two circles so that a segment of the same length as the radius of the small circle is marked off by CE in the larger circle. Construct a circle with center at O1 and passing through E. The side of the rectangle through E and parallel to the tangent is tangent to the smaller circle with center at O1 and passes through the center of the circle O2.

The concentric circle constructed in the small circle has radius equal to the difference of the two given radii. Now we can use Construction 2.4 to construct a tangent line from O2 to E. So the construction goes:

Step 1. Construct a concentric circle with center at the center of the larger of the two given circles, having a radius determined by the difference in the two radii.

Construction

Step 2. Construct the tangent from O2 to the circle of the difference of the radii.

Step 3. Construct the rectangle E(O2) DC.

Step 4. Construct the tangent line CD

Steps 5- 8. Repeat the process for the tangent line reflected in the line of centers.

Proof

The constructed rectangle with vertices at C and D will be on radii of the two circles, perpendicular to CD, and so CD is along the common tangent. For the case one with the two internal tangents, consider the tangent line with tangent points at F on O1 and H on O2. Construct a rectangle on FH.

Note that the tangent line from O2 to an expanded circle with center at O1 and radius equal to the sum of the two radii is along the side of the rectangle.

Therefore the construction of the tangent line through F and H would be very similar to our previous construction except that we use an auxiliary circle with a radius the sum of the two given radii rather than the difference.

The construction of the common internal tangent through G and K would be done in the same way.

ASSIGNMENT: (not really) I suggest you build GSP file with script tools for constructing these common tangents to two circles.

EXTRA: Consider constructing circles tangent to the two given circles. Explore the locus of the center of the tangent circles as the point of tangency on one of the circles is moved. Look for both internal tangent circles and external tangent circles.

Problem Set 2.3